# How do you find the exact values of cos(5pi/12) using the half angle formula?

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Alan P. Share
Mar 31, 2017

$\cos \left(\frac{5 \pi}{12}\right) = \frac{\sqrt{2 - \sqrt{3}}}{2}$

#### Explanation:

By the half angle formula:
$\textcolor{w h i t e}{\text{XXXX}}$$\cos \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \left(\theta\right)}{2}}$

If $\frac{\theta}{2} = \frac{5 \pi}{12}$
$\textcolor{w h i t e}{\text{XXXX}}$then $\theta = \frac{5 \pi}{6}$

Note that $\frac{5 \pi}{6}$ is a standard angle in quadrant 2 with a reference angle of $\frac{\pi}{6}$
so $\cos \left(\frac{5 \pi}{6}\right) = - \cos \left(\frac{\pi}{6}\right) = - \frac{\sqrt{3}}{2}$

Therefore
$\textcolor{w h i t e}{\text{XXXX}} \cos \left(\frac{5 \pi}{12}\right) = \pm \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}}$

$\textcolor{w h i t e}{\text{XXXXXXXXXXX}} = \pm \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}}$

$\textcolor{w h i t e}{\text{XXXXXXXXXXX}} = \pm \sqrt{\frac{2 - \sqrt{3}}{4}}$

$\textcolor{w h i t e}{\text{XXXXXXXXXXX}} = \pm \frac{\sqrt{2 - \sqrt{3}}}{2}$

Since $\frac{5 \pi}{12} < \frac{\pi}{2}$
$\textcolor{w h i t e}{\text{XXXX}}$$\frac{5 \pi}{12}$ is in quadrant 1
$\textcolor{w h i t e}{\text{XXXX}}$$\rightarrow \cos \left(\frac{5 \pi}{12}\right)$ is positive
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$(the negative solution is extraneous)

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